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According to Reynolds decomposition, velocity field is split into two time average and fluctuating parts:

$$u_{({\bf x},t)}=\overline u_{(\bf x)}+u'_{({\bf x},t)}$$

we know that time average of fluctuating part is zero $\overline u'=0$, but what if i take space average of fluctuating part?

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Do you want: 1) the space average of the fluctuating parts in the time-average decomposition or 2) the space average of the fluctuating part in a space average decomposition. I understand that you want 1) but looking at the other answer I'm not sure it is clear. –  aberration Aug 23 '13 at 15:54
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2 Answers

You may define the space average over some domain in exactly the same manner as the time average. Call it $[U(t)]$. Then a Reynolds-like decomposition is always possible: $$U(x,t) = [U(t)] + U'(x,t),$$ where $U'(x,t)$ is the fluctuation around the spatial average of the field.

Then by taking a spatial average of the above relation you obtain

$$[U(x,t)] = [[U(t)]] + [U'(x,t)] \rightarrow [U(t)] = [U(t)] + [U'(x,t)] \rightarrow [U'(x,t)] = 0$$

The spatial average of the fluctuation around the spatial average is zero.

Now it is not clear what is your purpose in asking the question because when you want to transform Navier Stokes by averaging methods, you either choose time averaging (RANS) or space averaging (LES) but not both. The relation here establishes the link between time and space averages but it is not very useful.

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LES (Large Eddy Simulation) is a filtering not an averaging (see en.wikipedia.org/wiki/Large_eddy_simulation) –  aberration Aug 22 '13 at 13:50
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Starting from the original statistical Reynolds decomposition where $\mathbf{u}$ is a random field, where $<...>_s$ denote the statistical average and where $\mathbf{u}^{(s)}$ is the random fluctuation field, it come the decomposition: \begin{equation} \mathbf{u}(\mathbf{x},t)=<u>_s(\mathbf{x},t)+\mathbf{u}^{(s)}(\mathbf{x},t) \end{equation} In your question it's look like you assume that the turbulence is stationary and that the ergodic hypothesis hold allowing to assimilate the statistical average to a time average and resulting of the decomposition: \begin{equation} \mathbf{u}(\mathbf{x},t)=<u>_t(\mathbf{x})+\mathbf{u}^{(t)}(\mathbf{x},t) \end{equation} where $<u>_s(\mathbf{x},t)=<u>_t(\mathbf{x})$ and the time average is defined by \begin{equation} <...>_t= \lim_{T \to +\infty} \frac{1}{T}\int_{t_0}^{t_0+T} ... dt \end{equation} Defining the spatial average over a bounded domain $\mathcal{D}$ of volume $V$: \begin{equation} <...>_{\mathbf{x}}=\frac{1}{V}\int_{\mathcal{D}} ... d\mathbf{x} \end{equation} For reply to your question, the space average of the previously defined fluctuating part is a random function of time $\mathbf{U}$ expressed \begin{equation} <\mathbf{u}^{(t)}(\mathbf{x},t)>_{\mathbf{x}}=\mathbf{U}(t)\in \mathbb{R}^3 \end{equation} Since $\mathcal{D}$ is bounded \begin{equation} lim_{T \to +\infty} \frac{1}{T}\int_{t_0}^{t_0+T} \frac{1}{V}\int_\mathcal{D}... d\mathbf{x} dt = \frac{1}{V} \int_\mathcal{D} ( lim_{T \to +\infty} \frac{1}{T}\int_{t_0}^{t_0+T} ... dt )d\mathbf{x} \end{equation} And it's result the property for $\mathbf{U}$: \begin{equation} <\mathbf{U}(t)>_t=\mathbf{0} \end{equation}

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